Differential equations substitution

**acevipa** · September 3rd 2010, 01:21 AM

Solve:

$2y'y''=1+(y')^2$ by letting $y'=v$

Does $y''=v'$

So does $2vv'=1+v^2$

**Ackbeet** · September 3rd 2010, 02:11 AM

Right. So what can you do now?

**acevipa** · September 3rd 2010, 02:26 AM

What is $v'$ with respect to?

**Ackbeet** · September 3rd 2010, 02:31 AM

The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.

**acevipa** · September 3rd 2010, 02:35 AM

Originally Posted by Ackbeet

The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.

So, $\dfrac{2v}{1+v^2} \ dv=dx$

$\displaystyle\int{\dfrac{2v}{1+v^2} \ dv}=\int{dx}$

$\ln|1+v^2|=x+C$

$1+v^2=Ce^x$

$v^2=Ce^x-1$

**Ackbeet** · September 3rd 2010, 02:40 AM

Not sure I would use the division symbol there. On the LHS, you have $dv/dx$ , originally. When you multiply both sides of

$\displaystyle{\frac{2v}{1+v^{2}}\,\frac{dv}{dx}=1}$ by $dx$ , what are you going to get?

[EDIT]: acevipa edited post # 5. It's correct now.

**acevipa** · September 3rd 2010, 02:43 AM

Didn't I do it correct?

**Ackbeet** · September 3rd 2010, 02:55 AM

Oh, I see. You edited your post. That looks good to me.

**acevipa** · September 3rd 2010, 03:00 AM

Originally Posted by Ackbeet

Oh, I see. You edited your post. That looks good to me.

Therefore, $(y')^2=Ce^x-1$

What do you do from here? Don't we have to find y?

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